= Prime-support bound in two-isogeny descent
{title2=$r\leq\nu(b)+\nu(a^2-4b)$}
For integral $a,b$ with $b(a^2-4b)\ne0$, every square class in the image of the <two-isogeny descent> map on $y^2=x(x^2+ax+b)$ has a signed square-free representative supported on <primes> dividing $b$. At a <prime> not dividing $b$, a positive <valuation> of $x$ must be even from the curve equation; a negative <valuation> is even because the cubic leading term determines $2v(y)=3v(x)$. Thus its image has at most $2^{\nu(b)+1}$ elements. Apply this also to the two-isogenous curve and use the <square-class index formula for two-isogeny descent> to obtain $r\leq\nu(b)+\nu(a^2-4b)$, where $\nu$ counts distinct <prime> divisors of the nonzero absolute value.
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